SQL中基于层次结构的重叠时间间隔合并

我试图解决一个问题，即我想要合并给定列id的重叠间隔，但我也想根据层次结构/优先级合并它们。我有每个间隔的开始时间和停止时间，每个间隔都有与其相关的层次结构/优先级

表中有以下列：

id, start_time, stop_time, priority

我能够解决我没有考虑优先权的问题，但我正在努力解决这个问题

Red colour: p1 (priority 1)
Blue Colour: p2 (priority 2)
Green colour: p3 (priority 3)

请注意，在下面的示例输入中，我们将有9行具有相同的id，而输出将有6行。请注意，对于某些id，可能只有一些优先级值或只有一个优先级值，解决方案应该考虑到这一点

预期投入和产出：

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这是一个复杂的问题。一个解决方案是找到岛屿的起点，并对起点进行累计。您可以通过查看没有重叠的位置来确定开始位置：

select id, priority, min(start_time), max(stop_time)
from (select t.*,
             countif(coalesce(prev_stop_time, stop_time) < stop_time) over (partition by id, priority order by start_time) as grp
      from (select t.*,
                   max(stop_time) over (partition by id, priority order by start_time rows between unbounded preceding and 1 preceding) as prev_stop_time
            from t
           ) t
      ) t
group by id, priority, grp;

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下面是BigQuery标准SQL

#standardSQL
WITH check_times AS (
  SELECT id, start_time AS time FROM `project.dataset.table` UNION DISTINCT
  SELECT id, stop_time AS time FROM `project.dataset.table` 
), distinct_intervals AS (
  SELECT id, time AS start_time, LEAD(time) OVER(PARTITION BY id ORDER BY time) stop_time
  FROM check_times
), deduped_intervals AS (
  SELECT a.id, a.start_time, a.stop_time, MIN(priority) priority
  FROM distinct_intervals a
  JOIN `project.dataset.table` b
  ON a.id = b.id 
  AND a.start_time BETWEEN b.start_time AND b.stop_time 
  AND a.stop_time BETWEEN b.start_time AND b.stop_time
  GROUP BY a.id, a.start_time, a.stop_time
), combined_intervals AS (
  SELECT id, MIN(start_time) start_time, MAX(stop_time) stop_time, ANY_VALUE(priority) priority 
  FROM (
    SELECT id, start_time, stop_time, priority, COUNTIF(flag) OVER(PARTITION BY id ORDER BY start_time) grp
    FROM (
      SELECT id, start_time, stop_time, priority, 
        start_time != IFNULL(LAG(stop_time) OVER(PARTITION BY id ORDER BY start_time), start_time) OR
        priority != IFNULL(LAG(priority) OVER(PARTITION BY id ORDER BY start_time), -1) flag
      FROM deduped_intervals
    )
  )
  GROUP BY id, grp
)
SELECT *
FROM combined_intervals
-- ORDER BY id, start_time

如果要应用于您问题中的样本数据-结果为

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您是否也可以共享一个解决方案，在该解决方案中，我们只基于id而不基于优先级列合并间隔

我只是稍微调整了一下上面的查询，以忽略优先级

#standardSQL
WITH check_times AS (
  SELECT id, start_time AS TIME FROM `project.dataset.table` UNION DISTINCT
  SELECT id, stop_time AS TIME FROM `project.dataset.table` 
), distinct_intervals AS (
  SELECT id, TIME AS start_time, LEAD(TIME) OVER(PARTITION BY id ORDER BY TIME) stop_time
  FROM check_times
), deduped_intervals AS (
  SELECT a.id, a.start_time, a.stop_time 
  FROM distinct_intervals a
  JOIN `project.dataset.table` b
  ON a.id = b.id 
  AND a.start_time BETWEEN b.start_time AND b.stop_time 
  AND a.stop_time BETWEEN b.start_time AND b.stop_time
  GROUP BY a.id, a.start_time, a.stop_time
), combined_intervals AS (
  SELECT id, MIN(start_time) start_time, MAX(stop_time) stop_time 
  FROM (
    SELECT id, start_time, stop_time, COUNTIF(flag) OVER(PARTITION BY id ORDER BY start_time) grp
    FROM (
      SELECT id, start_time, stop_time, 
        start_time != IFNULL(LAG(stop_time) OVER(PARTITION BY id ORDER BY start_time), start_time) flag
      FROM deduped_intervals
    )
  )
  GROUP BY id, grp
)
SELECT *
FROM combined_intervals
-- ORDER BY id, start_time   

结果

Row id  start_time  stop_time    
1   1   0           36   
2   1   41          47   

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MySQL还是BigQuery？他们完全不同。典型的缺口和孤岛问题。@GordonlinoffBigQuery。加了一些details@GordonLinoff另外，请不要告诉我，我需要为一个给定的id组合这些。我修改了我的答案以考虑到这一点。非常感谢@mikhail，非常好用。我将在更多的样本数据上测试它，以验证解决方案。当然。记住——答案基于你在问题中提出的逻辑和样本数据。所以，若你们要进入一个并没有涉及的案例，请发布新的问题。同时考虑表决，如果确认工作-回来接受：你也可以分享一个解决方案，我们合并间隔基于ID和没有优先列。我想看看它对我提出的解决方案的表现。我想比较一下运行时间，因为我处理的数据非常大。谢谢。将此添加到第二个解决方案中的我的answerHi@Mikhail合并中，而不考虑优先级。是否可以修改解决方案，以便我们在该时间段内再增加一列优先级，即优先级的最大值？因此，我们将有一个额外的优先级列，第10到36行的持续时间值为3，第2行的持续时间值为41到47。